What is the solution for y^{''}+20y^'+200=300 ?

The solution for y^{''}+20y^'+200=300 is y=c_{1}+c_{2}e^{-20t}+5t

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y^{''}+20y^'+200=300

y^{^{''}} + 20 y^{^{'}} + 200 = 300

y = c_{1} + c_{2} e^{- 20 t} + 5 t

Solution steps

y^{''} (t) + 20 y^{'} (t) + 200 = 300

Solve linear ODE:

y = c_{1} + c_{2} e^{- 20 t} + 5 t

y = c_{1} + c_{2} e^{- 20 t} + 5 t

\int_{0}^{N} e^{- s t} sin (2 t) d t

\frac{1}{1 + 4 t ^{2}}

\frac{4 x}{x + 1} (3.3)

y = t (ln (3 t))^{2}

\frac{d}{d x} (x^{3} + 2)

What is the solution for y^{''}+20y^'+200=300 ?
The solution for y^{''}+20y^'+200=300 is y=c_{1}+c_{2}e^{-20t}+5t